Kevin Venzke wrote:
As Blake pointed out, we can think of truncated votes as more or less equivalent to the same votes completed with random rankings. In that case, any margin of victory translates to a majority victory.
So, what do you make of my favorite scenario?:
49 A 24 B 27 C>B
Consider the ballots:
49 A 24 B 24 C>B
Then A (who beats B 49-48 and beats C 49-24) is the Condorcet winner. Note that this result was obtained by removing 3 ballots from the original set, and that there is no way to elect A by removing fewer ballots. Therefore, declaring A the winner is the equivalent of ignoring the wishes of 3 voters. Similarly, by eliminating 7 ballots (3 A's and 4 C>B's), then B (who beats A 47-46 and beats C 24-23) becomes the Condorcet winner. Or by eliminating 23 ballots (all of them A's), then C (who beats A 27-26 and beats B 27-24), becomes the Condorcet winner. In summary,
* Electing A is equivalent to discarding 3 ballots * Electing B is equivalent to discarding 7 ballots * Electing C is equivalent to discarding 23 ballots
Based on this argument, A should be elected.
I know you're asking, "What if the sincere preferences were 49 A>B>C + 24 B>C>A + 27 C>B>A? Then B would be the sincere Condorcet winner." But if this were the case, then the B>C>A voters would have been better off voting sincerely. So why didn't they?
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